Question: Given $w$ and $z$ are complex numbers such that $|w+z|=1$ and $|w^2+z^2|=14,$ find the smallest possible value of $|w^3+z^3|.$
Explanation: We try to express $w^3+z^3$ in terms of $w+z$ and $w^2+z^2.$ We have, by sum of cubes, \[w^3+z^3=(w+z)(w^2+z^2-wz),\]so we want now to express $wz$ in terms of $w+z$ and $w^2+z^2.$ To do that, we write $(w+z)^2 = w^2+z^2+2wz,$ from which it follows that $wz = \tfrac12 \left((w+z)^2 - (w^2+z^2)\right).$ Thus, \[\begin{aligned} w^3+z^3&=(w+z)(w^2+z^2-\tfrac12\left((w+z)^2-(w^2+z^2)\right)) \\ &= (w+z)\left(\tfrac32(w^2+z^2)-\tfrac12(w+z)^2\right). \end{aligned}\]Taking magnitudes of both sides, we have \[\begin{aligned} \left|w^3+z^3\right| &= \left| (w+z)\left(\tfrac32(w^2+z^2)-\tfrac12(w+z)^2\right) \right| \\ &=|w+z| \cdot \left|\tfrac32(w^2+z^2)-\tfrac12(w+z)^2\right|. \end{aligned}\]We are given that $|w+z| = 1,$ so \[|w^3+z^3| = \left|\tfrac32(w^2+z^2)-\tfrac12(w+z)^2\right|.\]We have $\left|\tfrac32(w^2+z^2)\right| = \tfrac32 \cdot 14 = 21$ and $\left|\tfrac12(w+z)^2\right| = \tfrac12 \cdot 1^2 = \tfrac12,$ so by the triangle inequality, \[|w^3+z^3| \ge \left| 21 - \tfrac12 \right| = \boxed{\tfrac{41}2}.\]